A geometry problem by Dexter Woo Teng Koon

Let $JC=1$ .

$\angle OJP=\arctan2=63.435^\circ$

$\angle KJC=180^\circ -2\times63.435^\circ=53.13^\circ$

$CK=\tan53.13^\circ=\dfrac43$

$k=\dfrac{CD}{CK}=\dfrac{4}{\frac43}=\boxed{3}$

Here is why it comes out exact:

$CK=\tan(180^\circ-2\arctan2)=-\dfrac{2\tan(\arctan2)}{1-\tan^2(\arctan2)}=-\dfrac{2\times2}{1-4}=\dfrac43$

Imagine a coordinate system with it's origin at the center of the circle. Then if $r$ is the radius of the circle, the equation that describes the circle is:
$x^{2}+y^{2}=r^{2}$

Let $I(x_{0},y_{0})$
$I$ is a point of the circle. Therefore:
$x_{0}^{2}+y_{0}^{2}=r^{2}$ ..... $(1)$

The equation of the tangent line of the circle at point $I$ is:
$x_{0}x+y_{0}y=r^{2}$

Let $J(x_{1},y_{1})$
It is obvious that $y_{1}=-r$
Also:
$4\times JC=BC\Rightarrow 4\times(r-x_{1})=2r\Rightarrow 4r-4x_{1}=2r\Rightarrow x_{1}=\frac{r}{2}$

$J$ is a point of the tangent line. Therefore:
$x_{0}\frac{r}{2}+y_{0}(-r)=r^{2}\Rightarrow x_{0}=2y_{0}+2$

$(1)\Rightarrow (2y_{0}+2)^{2}+y_{0}^2=r^{2}\Rightarrow 5y_{0}^2+8ry_{0}+3r^{2}=0\Rightarrow y_{0}=-a(declined)$ or $y_{0}=\frac{-3r}{5}$

$x_{0}=2y_{0}+2\Rightarrow x_{0}=\frac{4r}{5}$

Thus, the equation of the tangent line is:
$\frac{4r}{5}x-\frac{3r}{5}y=r^{2}\Rightarrow 4x-3y=5r$

For $x=r$ we have:
$4r-3y=5r\Rightarrow y=-\frac{r}{3}$

$KC=r-|y|\Rightarrow KC=\frac{2r}{3}$

$k\times KC=CD\Rightarrow k\frac{2r}{3}=2r\Rightarrow \boxed{k=3}$

Let the centre of the circle be $O$ , the point of tangency of the circle with $BC$ be $P$ . Without loss of generality, let $O(0,0)$ , $P(0,-1)$ , $C(1,-1)$ . It becomes obvious that $J(0.5,-1)$ . Now consider a line passing through $J$ with gradient $k$ . The line will have equation of the form $y=kx-0.5k-1$ We equate this with the equation of the circle $x^2=y^2=r^2$ to obtain: $x^2+(kx-0.5k-1)^2=1$ As the line must be tangent to the circle, the discriminant of this equation in $x$ must also be 0 to ensure that there is only one point of intersection. Thus $k=\frac{4}{3}$ . This results in $K(1,-\frac{2}{3})$ , yielding 3 as our desired answer.